Difference between revisions of "2022 AIME I Problems/Problem 9"

Revision as of 13:53, 7 April 2022

Problem

Ellina has twelve blocks, two each of red ( $\textbf{R}$ ), blue ( $\textbf{B}$ ), yellow ( $\textbf{Y}$ ), green ( $\textbf{G}$ ), orange ( $\textbf{O}$ ), and purple ( $\textbf{P}$ ). Call an arrangement of blocks $\textit{even}$ if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement $\textbf{R B B Y G G Y R O P P O}$ is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

Consider this position chart: $\textbf{1 2 3 4 5 6 7 8 9 10 11 12}$ Since there has to be an even number of spaces between each ball of the same color, spots $1$ , $3$ , $5$ , $7$ , $9$ , and $11$ contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is $6! \cdot 6!$ (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of $\frac{12!}{(2!)^6}$ possible arrangements, so the probability is: $\frac{6!\cdot6!}{\frac{12!}{(2!)^6}} = \frac{6!\cdot2^6}{7\cdot8\cdot9\cdot10\cdot11\cdot12} = \frac{2^4}{7\cdot11\cdot3} = \frac{16}{231},$ which is in simplest form. So, $m + n = 16 + 231 = \boxed{247}$ .

~Oxymoronic15

Solution 2

We can simply use constructive counting. First, let us place the red balls; choose the first slot in $12$ ways, and the second in $6$ ways, because the number is cut in half due to the condition in the problem. This gives $12 \cdot 6$ ways to place the blue balls. Similarly, there are $10 \cdot 5$ ways to place the blue balls, and so on, until there are $2 \cdot 1$ ways to place the purple balls. Thus, the probability is $\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{231}$ , and the desired answer extraction is $16+231=\boxed{\textbf{247}}$ .

~A1001

Video Solution (Mathematical Dexterity)

https://www.youtube.com/watch?v=dkoF7StwtrM

Video Solution (Power of Logic)

https://youtu.be/AF6TOG7MSwA

@@ Line 12: / Line 12: @@
 ==Solution 2==
-We can simply use constructive counting. First, let us place the red balls; choose the first slot in <math>12</math> ways, and the second in <math>6</math> ways, because the number is cut in half due to the condition in the problem. This gives <math>12 \cdot 6</math> ways to place the blue balls. Similarly, there are <math>10 \cdot 5</math> ways to place the blue balls, and so on, until there are <math>2 \cdot 1</math> ways to place the purple balls. Thus, the probability is <cmath>\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{123}</cmath>, and the desired answer extraction is <math>16+123=\boxed{\textbf{139}}</math>.
+We can simply use constructive counting. First, let us place the red balls; choose the first slot in <math>12</math> ways, and the second in <math>6</math> ways, because the number is cut in half due to the condition in the problem. This gives <math>12 \cdot 6</math> ways to place the blue balls. Similarly, there are <math>10 \cdot 5</math> ways to place the blue balls, and so on, until there are <math>2 \cdot 1</math> ways to place the purple balls. Thus, the probability is <cmath>\frac{12 \cdot 6 \cdot 10 \cdot 5 \cdot 8 \cdot 4 \cdot 6 \cdot 3 \cdot 4 \cdot 2 \cdot 2 \cdot 1}{12!}=\frac{16}{231}</cmath>, and the desired answer extraction is <math>16+231=\boxed{\textbf{247}}</math>.
 ~A1001

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search